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JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties

\(\mathrm{CO}_{2}\) gas is bubbled through water during a soft drink manufacturing process at \(298\, \mathrm{~K}\). If \(\mathrm{CO}_{2}\) exerts a partial pressure of \(0.835\) bar then \(x \,\mathrm{~m} \,\mathrm{~mol}\) of \(\mathrm{CO}_{2}\) would dissolve in \(0.9\,\mathrm{~L}\) of water. The value of \(x\) is \(.....\) (Nearest integer) (Henry's law constant for \(\mathrm{CO}_{2}\) at \(298\, \mathrm{~K}\) is \(1.67 \times 10^{3}\) \(bar\))

  1. A \(50\)
  2. B \(25\)
  3. C \(55\)
  4. D \(35\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(25\)

Step-by-step Solution

Detailed explanation

From Henry's Law \(0.835=1.67 \times 10^{3} \times \frac{\mathrm{n}\left(\mathrm{CO}_{2}\right)}{\frac{0.9 \times 1000}{18}}\) \(\mathrm{n}\left(\mathrm{CO}_{2}\right)=0.025\) Millimoles of \(\mathrm{CO}_{2}=0.025 \times 1000=25\)
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